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JEE Mains Previous Year Questions 
(2021-2024): Structure of Atom 
2024 
Q1: The ionization energy of sodium in ???? ?????? -?? , if electromagnetic radiation of wavelength ?????? ???? 
is just sufficient to ionize sodium atom is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 494 
E =
1240
?? (nm)
eV
 =
1240
242
eV
 = 5.12eV
 = 5.12 × 1.6 × 10
-19
 = 8.198 × 10
-19
 J/ atom 
 = 494 kJ/mol
 
Q2: Number of spectral lines obtained in ????
+
spectra, when an electron makes transition from fifth 
excited state to first excited state will be    [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: 10 
5
th 
 excited state ? ?? 1
= 6
1
st 
 excited state ? ?? 2
= 2
??? = ?? 1
- ?? 2
= 6 - 2 = 4
 
Maximum number of spectral lines 
=
?n(?n + 1)
2
=
4(4 + 1)
2
= 10 
Q3: The number of electrons present in all the completely filled subshells having ?? = ?? and ?? = +
?? ?? is 
(Where ?? = principal quantum number and ?? = spin quantum number)     
[JEE Main 2024 (Online) 27th January Morning Shift] 
Ans: 16 
To determine the number of electrons with a spin quantum number of ?? = +
1
2
 in all completely filled 
subshells with principal quantum number ?? = 4, we must first identify the subshells in the ?? = 4 shell 
and then calculate the electrons with the specified spin. 
The ?? = 4 shell has the following subshells and their capacity for electrons: 
4?? subshell can hold 2 electrons 
4?? subshell can hold 6 electrons 
4?? subshell can hold 10 electrons 
4?? subshell can hold 14 electrons 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Structure of Atom 
2024 
Q1: The ionization energy of sodium in ???? ?????? -?? , if electromagnetic radiation of wavelength ?????? ???? 
is just sufficient to ionize sodium atom is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 494 
E =
1240
?? (nm)
eV
 =
1240
242
eV
 = 5.12eV
 = 5.12 × 1.6 × 10
-19
 = 8.198 × 10
-19
 J/ atom 
 = 494 kJ/mol
 
Q2: Number of spectral lines obtained in ????
+
spectra, when an electron makes transition from fifth 
excited state to first excited state will be    [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: 10 
5
th 
 excited state ? ?? 1
= 6
1
st 
 excited state ? ?? 2
= 2
??? = ?? 1
- ?? 2
= 6 - 2 = 4
 
Maximum number of spectral lines 
=
?n(?n + 1)
2
=
4(4 + 1)
2
= 10 
Q3: The number of electrons present in all the completely filled subshells having ?? = ?? and ?? = +
?? ?? is 
(Where ?? = principal quantum number and ?? = spin quantum number)     
[JEE Main 2024 (Online) 27th January Morning Shift] 
Ans: 16 
To determine the number of electrons with a spin quantum number of ?? = +
1
2
 in all completely filled 
subshells with principal quantum number ?? = 4, we must first identify the subshells in the ?? = 4 shell 
and then calculate the electrons with the specified spin. 
The ?? = 4 shell has the following subshells and their capacity for electrons: 
4?? subshell can hold 2 electrons 
4?? subshell can hold 6 electrons 
4?? subshell can hold 10 electrons 
4?? subshell can hold 14 electrons 
In quantum mechanics, each orbital within these subshells can hold 2 electrons, each with opposite 
spins ( +
1
2
 and -
1
2
 ). Thus, in a completely filled subshell, half of the electrons will have a spin quantum 
number of +
1
2
 : 
In 4?? , 1 out of 2 electrons will have ?? = +
1
2
 
In 4?? , 3 out of 6 electrons will have ?? = +
1
2
 
In 4?? , 5 out of 10 electrons will have ?? = +
1
2
 
In 4?? , 7 out of 14 electrons will have ?? = +
1
2
 
Adding these together: 1 + 3 + 5 + 7 = 16 electrons with ?? = +
1
2
 in all completely filled subshells with 
?? = 4. 
Q4: The number of radial node/s for ?? ?? orbital is : 
A. 3 
B. 2 
C. 1 
D. 4   [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
The number of radial nodes in an orbital is given by the formula: 
Number of radial nodes = ?? - ?? - 1 
where ?? is the principal quantum number, ?? is the azimuthal quantum number also known as the 
angular momentum quantum number. 
For a 3?? orbital, 
?? = 3 (since it is the 3rd energy level), 
?? = 1 (since ?? orbitals correspond to ?? = 1 ). 
Let's use the formula to determine the number of radial nodes: 
Number of radial nodes = 3 - 1 - 1 = 1 
So, the correct answer is: Option C: 1 
Q5: In case of isoelectronic species the size of ?? -
, ???? and ????
+
is affected by: 
A. Nuclear charge (?? ) 
B. None of the factors because their size is the same 
C. Electron-electron interaction in the outer orbitals 
D. Principal quantum number (n)   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
Option A: Nuclear charge (z) is the correct answer. 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Structure of Atom 
2024 
Q1: The ionization energy of sodium in ???? ?????? -?? , if electromagnetic radiation of wavelength ?????? ???? 
is just sufficient to ionize sodium atom is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 494 
E =
1240
?? (nm)
eV
 =
1240
242
eV
 = 5.12eV
 = 5.12 × 1.6 × 10
-19
 = 8.198 × 10
-19
 J/ atom 
 = 494 kJ/mol
 
Q2: Number of spectral lines obtained in ????
+
spectra, when an electron makes transition from fifth 
excited state to first excited state will be    [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: 10 
5
th 
 excited state ? ?? 1
= 6
1
st 
 excited state ? ?? 2
= 2
??? = ?? 1
- ?? 2
= 6 - 2 = 4
 
Maximum number of spectral lines 
=
?n(?n + 1)
2
=
4(4 + 1)
2
= 10 
Q3: The number of electrons present in all the completely filled subshells having ?? = ?? and ?? = +
?? ?? is 
(Where ?? = principal quantum number and ?? = spin quantum number)     
[JEE Main 2024 (Online) 27th January Morning Shift] 
Ans: 16 
To determine the number of electrons with a spin quantum number of ?? = +
1
2
 in all completely filled 
subshells with principal quantum number ?? = 4, we must first identify the subshells in the ?? = 4 shell 
and then calculate the electrons with the specified spin. 
The ?? = 4 shell has the following subshells and their capacity for electrons: 
4?? subshell can hold 2 electrons 
4?? subshell can hold 6 electrons 
4?? subshell can hold 10 electrons 
4?? subshell can hold 14 electrons 
In quantum mechanics, each orbital within these subshells can hold 2 electrons, each with opposite 
spins ( +
1
2
 and -
1
2
 ). Thus, in a completely filled subshell, half of the electrons will have a spin quantum 
number of +
1
2
 : 
In 4?? , 1 out of 2 electrons will have ?? = +
1
2
 
In 4?? , 3 out of 6 electrons will have ?? = +
1
2
 
In 4?? , 5 out of 10 electrons will have ?? = +
1
2
 
In 4?? , 7 out of 14 electrons will have ?? = +
1
2
 
Adding these together: 1 + 3 + 5 + 7 = 16 electrons with ?? = +
1
2
 in all completely filled subshells with 
?? = 4. 
Q4: The number of radial node/s for ?? ?? orbital is : 
A. 3 
B. 2 
C. 1 
D. 4   [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
The number of radial nodes in an orbital is given by the formula: 
Number of radial nodes = ?? - ?? - 1 
where ?? is the principal quantum number, ?? is the azimuthal quantum number also known as the 
angular momentum quantum number. 
For a 3?? orbital, 
?? = 3 (since it is the 3rd energy level), 
?? = 1 (since ?? orbitals correspond to ?? = 1 ). 
Let's use the formula to determine the number of radial nodes: 
Number of radial nodes = 3 - 1 - 1 = 1 
So, the correct answer is: Option C: 1 
Q5: In case of isoelectronic species the size of ?? -
, ???? and ????
+
is affected by: 
A. Nuclear charge (?? ) 
B. None of the factors because their size is the same 
C. Electron-electron interaction in the outer orbitals 
D. Principal quantum number (n)   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
Option A: Nuclear charge (z) is the correct answer. 
Isoelectronic species are atoms and ions that have the same number of electrons. When comparing the 
sizes of isoelectronic species, the major factor that affects their size is the nuclear charge ( z), which is 
the charge of the nucleus or the number of protons in the nucleus. Here's why: 
The more protons there are in the nucleus (higher nuclear charge), the more positively charged the 
nucleus is, and thus the stronger it can attract the electrons towards it. This results in a smaller radius 
for the species. Conversely, if there are fewer protons in the nucleus, the attraction to the electrons is 
weaker, leading to a larger radius. Let's compare the species given: 
F
-
(fluoride ion) has 9 protons in the nucleus and 10 electrons. 
Ne (neon atom) has 10 protons in the nucleus and 10 electrons. 
Na
+
(sodium ion) has 11 protons in the nucleus and 10 electrons. 
Despite having the same number of electrons, the Na
+
ion will be the smallest among the three due to 
the higher nuclear charge of 11 protons, which attracts the same number of electrons more strongly 
than Ne with 10 protons and F
-
with 9 protons. Hence, Na
+
has a lower atomic radius compared to Ne, 
which in turn is smaller than F
-
. 
Considering Option B (size being the same), Option C (electron-electron interaction), and Option D 
(principal quantum number), none of these factors are as significant as the difference in nuclear charge 
for isoelectronic species. 
The principal quantum number (?? ) is the same for all these isoelectronic species since they all have 
their outer electrons in the second energy level (?? = 2) in the ground state. Electron-electron 
interactions do contribute to energy levels and could affect the size slightly, but the primary and 
determining factor for isoelectronic species remains the nuclear charge, as it directly influences the 
effective nuclear charge felt by the electrons. Therefore, Option A is the best answer. 
Q6: According to the wave-particle duality of matter by de-Broglie, which of the following graph plot 
presents most appropriate relationship between wavelength of electron (?? ) and momentum of 
electron (?? ) ?   [JEE Main 2024 (Online) 1st February Morning Shift] 
(a)  
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Structure of Atom 
2024 
Q1: The ionization energy of sodium in ???? ?????? -?? , if electromagnetic radiation of wavelength ?????? ???? 
is just sufficient to ionize sodium atom is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 494 
E =
1240
?? (nm)
eV
 =
1240
242
eV
 = 5.12eV
 = 5.12 × 1.6 × 10
-19
 = 8.198 × 10
-19
 J/ atom 
 = 494 kJ/mol
 
Q2: Number of spectral lines obtained in ????
+
spectra, when an electron makes transition from fifth 
excited state to first excited state will be    [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: 10 
5
th 
 excited state ? ?? 1
= 6
1
st 
 excited state ? ?? 2
= 2
??? = ?? 1
- ?? 2
= 6 - 2 = 4
 
Maximum number of spectral lines 
=
?n(?n + 1)
2
=
4(4 + 1)
2
= 10 
Q3: The number of electrons present in all the completely filled subshells having ?? = ?? and ?? = +
?? ?? is 
(Where ?? = principal quantum number and ?? = spin quantum number)     
[JEE Main 2024 (Online) 27th January Morning Shift] 
Ans: 16 
To determine the number of electrons with a spin quantum number of ?? = +
1
2
 in all completely filled 
subshells with principal quantum number ?? = 4, we must first identify the subshells in the ?? = 4 shell 
and then calculate the electrons with the specified spin. 
The ?? = 4 shell has the following subshells and their capacity for electrons: 
4?? subshell can hold 2 electrons 
4?? subshell can hold 6 electrons 
4?? subshell can hold 10 electrons 
4?? subshell can hold 14 electrons 
In quantum mechanics, each orbital within these subshells can hold 2 electrons, each with opposite 
spins ( +
1
2
 and -
1
2
 ). Thus, in a completely filled subshell, half of the electrons will have a spin quantum 
number of +
1
2
 : 
In 4?? , 1 out of 2 electrons will have ?? = +
1
2
 
In 4?? , 3 out of 6 electrons will have ?? = +
1
2
 
In 4?? , 5 out of 10 electrons will have ?? = +
1
2
 
In 4?? , 7 out of 14 electrons will have ?? = +
1
2
 
Adding these together: 1 + 3 + 5 + 7 = 16 electrons with ?? = +
1
2
 in all completely filled subshells with 
?? = 4. 
Q4: The number of radial node/s for ?? ?? orbital is : 
A. 3 
B. 2 
C. 1 
D. 4   [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
The number of radial nodes in an orbital is given by the formula: 
Number of radial nodes = ?? - ?? - 1 
where ?? is the principal quantum number, ?? is the azimuthal quantum number also known as the 
angular momentum quantum number. 
For a 3?? orbital, 
?? = 3 (since it is the 3rd energy level), 
?? = 1 (since ?? orbitals correspond to ?? = 1 ). 
Let's use the formula to determine the number of radial nodes: 
Number of radial nodes = 3 - 1 - 1 = 1 
So, the correct answer is: Option C: 1 
Q5: In case of isoelectronic species the size of ?? -
, ???? and ????
+
is affected by: 
A. Nuclear charge (?? ) 
B. None of the factors because their size is the same 
C. Electron-electron interaction in the outer orbitals 
D. Principal quantum number (n)   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
Option A: Nuclear charge (z) is the correct answer. 
Isoelectronic species are atoms and ions that have the same number of electrons. When comparing the 
sizes of isoelectronic species, the major factor that affects their size is the nuclear charge ( z), which is 
the charge of the nucleus or the number of protons in the nucleus. Here's why: 
The more protons there are in the nucleus (higher nuclear charge), the more positively charged the 
nucleus is, and thus the stronger it can attract the electrons towards it. This results in a smaller radius 
for the species. Conversely, if there are fewer protons in the nucleus, the attraction to the electrons is 
weaker, leading to a larger radius. Let's compare the species given: 
F
-
(fluoride ion) has 9 protons in the nucleus and 10 electrons. 
Ne (neon atom) has 10 protons in the nucleus and 10 electrons. 
Na
+
(sodium ion) has 11 protons in the nucleus and 10 electrons. 
Despite having the same number of electrons, the Na
+
ion will be the smallest among the three due to 
the higher nuclear charge of 11 protons, which attracts the same number of electrons more strongly 
than Ne with 10 protons and F
-
with 9 protons. Hence, Na
+
has a lower atomic radius compared to Ne, 
which in turn is smaller than F
-
. 
Considering Option B (size being the same), Option C (electron-electron interaction), and Option D 
(principal quantum number), none of these factors are as significant as the difference in nuclear charge 
for isoelectronic species. 
The principal quantum number (?? ) is the same for all these isoelectronic species since they all have 
their outer electrons in the second energy level (?? = 2) in the ground state. Electron-electron 
interactions do contribute to energy levels and could affect the size slightly, but the primary and 
determining factor for isoelectronic species remains the nuclear charge, as it directly influences the 
effective nuclear charge felt by the electrons. Therefore, Option A is the best answer. 
Q6: According to the wave-particle duality of matter by de-Broglie, which of the following graph plot 
presents most appropriate relationship between wavelength of electron (?? ) and momentum of 
electron (?? ) ?   [JEE Main 2024 (Online) 1st February Morning Shift] 
(a)  
(b)  
(c)  
(d)  
Ans: (d) 
 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Structure of Atom 
2024 
Q1: The ionization energy of sodium in ???? ?????? -?? , if electromagnetic radiation of wavelength ?????? ???? 
is just sufficient to ionize sodium atom is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 494 
E =
1240
?? (nm)
eV
 =
1240
242
eV
 = 5.12eV
 = 5.12 × 1.6 × 10
-19
 = 8.198 × 10
-19
 J/ atom 
 = 494 kJ/mol
 
Q2: Number of spectral lines obtained in ????
+
spectra, when an electron makes transition from fifth 
excited state to first excited state will be    [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: 10 
5
th 
 excited state ? ?? 1
= 6
1
st 
 excited state ? ?? 2
= 2
??? = ?? 1
- ?? 2
= 6 - 2 = 4
 
Maximum number of spectral lines 
=
?n(?n + 1)
2
=
4(4 + 1)
2
= 10 
Q3: The number of electrons present in all the completely filled subshells having ?? = ?? and ?? = +
?? ?? is 
(Where ?? = principal quantum number and ?? = spin quantum number)     
[JEE Main 2024 (Online) 27th January Morning Shift] 
Ans: 16 
To determine the number of electrons with a spin quantum number of ?? = +
1
2
 in all completely filled 
subshells with principal quantum number ?? = 4, we must first identify the subshells in the ?? = 4 shell 
and then calculate the electrons with the specified spin. 
The ?? = 4 shell has the following subshells and their capacity for electrons: 
4?? subshell can hold 2 electrons 
4?? subshell can hold 6 electrons 
4?? subshell can hold 10 electrons 
4?? subshell can hold 14 electrons 
In quantum mechanics, each orbital within these subshells can hold 2 electrons, each with opposite 
spins ( +
1
2
 and -
1
2
 ). Thus, in a completely filled subshell, half of the electrons will have a spin quantum 
number of +
1
2
 : 
In 4?? , 1 out of 2 electrons will have ?? = +
1
2
 
In 4?? , 3 out of 6 electrons will have ?? = +
1
2
 
In 4?? , 5 out of 10 electrons will have ?? = +
1
2
 
In 4?? , 7 out of 14 electrons will have ?? = +
1
2
 
Adding these together: 1 + 3 + 5 + 7 = 16 electrons with ?? = +
1
2
 in all completely filled subshells with 
?? = 4. 
Q4: The number of radial node/s for ?? ?? orbital is : 
A. 3 
B. 2 
C. 1 
D. 4   [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
The number of radial nodes in an orbital is given by the formula: 
Number of radial nodes = ?? - ?? - 1 
where ?? is the principal quantum number, ?? is the azimuthal quantum number also known as the 
angular momentum quantum number. 
For a 3?? orbital, 
?? = 3 (since it is the 3rd energy level), 
?? = 1 (since ?? orbitals correspond to ?? = 1 ). 
Let's use the formula to determine the number of radial nodes: 
Number of radial nodes = 3 - 1 - 1 = 1 
So, the correct answer is: Option C: 1 
Q5: In case of isoelectronic species the size of ?? -
, ???? and ????
+
is affected by: 
A. Nuclear charge (?? ) 
B. None of the factors because their size is the same 
C. Electron-electron interaction in the outer orbitals 
D. Principal quantum number (n)   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
Option A: Nuclear charge (z) is the correct answer. 
Isoelectronic species are atoms and ions that have the same number of electrons. When comparing the 
sizes of isoelectronic species, the major factor that affects their size is the nuclear charge ( z), which is 
the charge of the nucleus or the number of protons in the nucleus. Here's why: 
The more protons there are in the nucleus (higher nuclear charge), the more positively charged the 
nucleus is, and thus the stronger it can attract the electrons towards it. This results in a smaller radius 
for the species. Conversely, if there are fewer protons in the nucleus, the attraction to the electrons is 
weaker, leading to a larger radius. Let's compare the species given: 
F
-
(fluoride ion) has 9 protons in the nucleus and 10 electrons. 
Ne (neon atom) has 10 protons in the nucleus and 10 electrons. 
Na
+
(sodium ion) has 11 protons in the nucleus and 10 electrons. 
Despite having the same number of electrons, the Na
+
ion will be the smallest among the three due to 
the higher nuclear charge of 11 protons, which attracts the same number of electrons more strongly 
than Ne with 10 protons and F
-
with 9 protons. Hence, Na
+
has a lower atomic radius compared to Ne, 
which in turn is smaller than F
-
. 
Considering Option B (size being the same), Option C (electron-electron interaction), and Option D 
(principal quantum number), none of these factors are as significant as the difference in nuclear charge 
for isoelectronic species. 
The principal quantum number (?? ) is the same for all these isoelectronic species since they all have 
their outer electrons in the second energy level (?? = 2) in the ground state. Electron-electron 
interactions do contribute to energy levels and could affect the size slightly, but the primary and 
determining factor for isoelectronic species remains the nuclear charge, as it directly influences the 
effective nuclear charge felt by the electrons. Therefore, Option A is the best answer. 
Q6: According to the wave-particle duality of matter by de-Broglie, which of the following graph plot 
presents most appropriate relationship between wavelength of electron (?? ) and momentum of 
electron (?? ) ?   [JEE Main 2024 (Online) 1st February Morning Shift] 
(a)  
(b)  
(c)  
(d)  
Ans: (d) 
 
?? =
h
?? ?? ?
1
?? 
So, graph between ?? and ?? is rectangular hyperbola. 
Q7: The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19)  
A. ?? = ?? , ?? = ?? , ?? = ?? , ?? = +
?? ?? 
B. ?? = ?? , ?? = ?? , ?? = ?? , ?? = +
?? ?? 
C. ?? = ?? , ?? = ?? , ?? = ?? , ?? = +
?? ?? 
D. ?? = ?? , ?? = ?? , ?? = -?? , ?? = +
?? ??   are    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
 
19
 K 1 s
2
, 2 s
2
, 2p
6
, 3 s
2
, 3p
6
, 4 s
1
. 
Outermost orbital of potassium is 4 s orbital 
n = 4,1 = 0, m
l
= 0, s = ±
1
2
 
Q8: Given below are two statements : 
Statement (I) : The orbitals having same energy are called as degenerate orbitals. 
Statement (II) : In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Statement I is true but Statement II is false 
B. Statement I is false but Statement II is true 
C. Both Statement I and Statement II are false 
D. Both Statement I and Statement II are true   [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: (a) 
For single electron species the energy depends upon principal quantum number 'n' only. So, statement II 
is false. 
Statement I is correct definition of degenerate orbitals. 
 
 
 
 
 
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Atomic Structure: JEE Mains Previous Year Questions (2021-2024) | Chemistry for JEE Main & Advanced

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